Events & Promotions
|
|
GMAT Club Daily Prep
Thank you for using the timer - this advanced tool can estimate your performance and suggest more practice questions. We have subscribed you to Daily Prep Questions via email.
Customized
for You
Track
Your Progress
Practice
Pays
Not interested in getting valuable practice questions and articles delivered to your email? No problem, unsubscribe here.
Nov 2007:30 AM PST
-08:30 AM PST
Learn what truly sets the UC Riverside MBA apart and how it helps in your professional growth
Nov 2211:00 AM IST
-01:00 PM IST
Do RC/MSR passages scare you? e-GMAT is conducting a masterclass to help you learn – Learn effective reading strategies Tackle difficult RC & MSR with confidence Excel in timed test environment- Nov 23
11:00 AM IST
-01:00 PM IST
Attend this free GMAT Algebra Webinar and learn how to master the most challenging Inequalities and Absolute Value problems with ease. 
Nov 2510:00 AM EST
-11:00 AM EST
Prefer video-based learning? The Target Test Prep OnDemand course is a one-of-a-kind video masterclass featuring 400 hours of lecture-style teaching by Scott Woodbury-Stewart, founder of Target Test Prep and one of the most accomplished GMAT instructors.
12 Feb 2021, 03:00
Kudos
Bookmarks
B
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (02:41) correct
67%
(02:35)
wrong
based on 51
sessions
History
Date
Time
Result
Not Attempted Yet
In an election for the managing committee of a reputed club, the number of candidates contesting elections exceeds the number of members to be elected by r (r > 0). If a voter can vote in 967 different ways to elect managing committee by voting at least 1 of them & can vote in 55 different ways to elect (r – 1) candidates by voting in the same manner. What is the number of candidates contesting the election & the number of candidates losing the elections?
A. 8and 3
B. 10 and 3
C. 12 and 3
D. 10 and 4
E. 10 and 5
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
A. 8and 3
B. 10 and 3
C. 12 and 3
D. 10 and 4
E. 10 and 5
Are You Up For the Challenge: 700 Level Questions: 700 Level Questions
12 Feb 2021, 20:30
Kudos
Bookmarks
Bunuel
Let us simplify the question by breaking it into variables.
Let the total number of the candidates be n.
The number of members to be elected =n-r
Thus a voter can vote for 1 of them or 2 of them till n-r in 967 ways.
That is nC1+nC2+nC3+....nC(n-r)=967 ways
The above sum should immediately get you to the formula
nC0+nC1+....nCn=2^n
So we have to get above formula into this form.
Add nC0 to both sides.
nC0+ nC1+nC2+nC3+....nC(n-r)=967+nC0=967+1=968....(I)
Now, let us take the second part.
A voter can vote in 55 different ways to elect (r – 1) candidates by voting in the same manner
So nC1+nC2+...nC(r-1)=55
nC0+nC1+nC2+.....nC(r-1)=55+nC0=55+1=56.....(II)
The other thing one should know is that nC0=nCn and nC1=nC(n-1)
Replace these values in (II)
nCn+nC(n-1)+....nC(n-(r-1))=56
Add equation (I) to above equation
\(nC0+nC1+....nC(n-r)+nC(n-(r-1))+.....nC(n-(r-r))=968+56=1024=2^n.......n=10\)
So total candidates are 10.
10C1+10C2+...10C(r-1)=55
10+45+..=55
So, r-1=2.......r=3
Answer : 10 and 3
B
26 Dec 2024, 07:35
Kudos
Bookmarks
Bunuel
Question language can be a little confusing at first glance.
Required knowledge -
\(nC0 + nC1 + nC2 + nC3 + .... + nCn = 2^n\)
Let the number of candidates be n and elected candidates be e, then n - e = r
I tried it with intuition and taking answer choices as reference.
967 => This number is closest to 1024, which is \(2^{10}\)
We have 3 answer choices (B, D, E) with n = 10, so with probability in our favor it's best that we assume this to be true and try calculating e or r.
\(10C0 + 10C1 + nC2 + nC3 + .... + 10C10 = 2^{10}\)
We now need to exclude 10C0, and few other combinations from the end to equate to 967
\(10C1 + 10C2 + 10C3 + .... + 10Ce = 1024 - 10C0 - 10C10 - 10C9 - .... - 10C(e+1)\) --- (1)
As we can see r is a relatively small number (3, 4, 5) from the option choice, we just need to subtract limited combinations
1024 - 10C0 - 10C10 - 10C9 - 10C8 --- (2)
1024 - 1 - 1 - 10 - 45
967 => Required total combinations
From 1 and 2, e + 1 = 8 => e = 7
r = n - e = 10 - 7 = 3
There's high probability that this is the correct answer choice. If time permits and want to verify the same from the additional condition - 55 different ways to elect (r – 1) candidates
which means, 10C1 + 10C2 = 55?
This can be easily verified and we can confirm that we have selected the right choice.
Answer: B
